Why would the two areas not shrinking together cause the total charge to become infinite?

Question

I am currently studying Principles of Optics: Electromagnetic Theory of Propagation, Interference and Diffraction of Light, 7th edition, by Max Born and Emil Wolf. Chapter 1.1.3 Boundary conditions at a surface of discontinuity says the following:

Let us replace the sharp discontinuity surface $T$ by a thin transition layer within which $\epsilon$ and $\mu$ vary rapidly but continuously from their values near $T$ on one side to their value near $T$ on the other. Within this layer we construct a small near-cylinder, bounded by a stockade of normals to $T$; roofed and floored by small area $\delta A_1$ and $\delta A_2$ on each side of $T$, at constant distance from it, measured along their common normal (Fig. 1.1). Since $\mathbf{B}$ and its derivatives may be assumed to be continuous throughout this cylinder, we may apply Gauss's theorem to the integral of $\text{div} \ \mathbf{B}$ taken throughout the volume of the cylinder and obtain, from (4), $$\int \text{div} \ \mathbf{B} \ dV = \int \mathbf{B} \cdot \mathbf{n} \ dS = 0; \tag{12}$$ the second integral is taken over the surface of the cylinder, and $\mathbf{n}$ is the unit outward normal. Since the area $\delta A_1$ and $\delta A_2$ are assumed to be small, $\mathbf{B}$ may be considered to have constants values $\mathbf{B}^{(1)}$ and $\mathbf{B}^{(2)}$ on $\delta A_1$ and $\delta A_2$, and (12) may then be replaced by $$\mathbf{B}^{(1)} \cdot \mathbf{n}_1 \delta A_1 + \mathbf{B}^{(2)} \cdot \mathbf{n}_2 \delta A_2 + \text{contribution from walls} = 0 \tag{13}$$ If the height $\delta h$ of the cylinder decreases towards zero, the transition layer shrinks into the surface and the contribution from the walls of the cylinder tends to zero, provided that there is no surface flux of magnetic induction. Such flux never occurs, and consequently in the limit, $$(\mathbf{B}^{(1)} \cdot \mathbf{n}_1 + \mathbf{B}^{(2)}\cdot \mathbf{n}_2) \delta A = 0, \tag{14}$$ $\delta A$ being the area in which the cylinder intersects $T$. If $\mathbf{n}_{12}$ is the unit normal pointing from the first into the second medium, then $\mathbf{n}_1 = - \mathbf{n}_{12}$, $\mathbf{n} = \mathbf{n}_{12}$ and (14) gives $$\mathbf{n}_{12} \cdot (\mathbf{B}^{(2)} - \mathbf{B}^{(1)}) = 0, \tag{15}$$ i.e. the normal component of the magnetic induction is continuous across the surface of discontinuity. The electric displacement $\mathbf{D}$ may be treated in a similar way, but there will be an additional term if charges are present. In place of (12) we now have from (3) $$\int \text{div} \ \mathbf{D} \ dV = \int \mathbf{D} \cdot \mathbf{n} \ dS = 4\pi \int \rho \ dV \tag{16}$$ As the areas $\delta A_1$ and $\delta A_2$ shrink together, the total charge remains finite, so that the volume density becomes infinite.

$\epsilon$ is the dielectric constant (or permittivity).
$\mu$ is the magnetic permeability.
$\mathbf{B}$ is the magnetic induction.
(4) is the relation $\text{div} \ \mathbf{B} = 0$.
$\rho$ is the electric charge density.
And (16) uses the relation $\text{div} \ \mathbf{D} = 4 \pi \rho$.

So it is said that, as the areas $\delta A_1$ and $\delta A_2$ shrink together, the total charge remains finite. But why would the two areas not shrinking together cause the total charge to become infinite? I don't understand this part. Furthermore, why does this then imply that the volume density becomes infinite?

I would greatly appreciate it if people would please take the time to clarify this.

Answer 1

So it is said that, as the areas $\delta A_1$ and $\delta A_2$ shrink together, the total charge remains finite. But why would the two areas not shrinking together cause the total charge to become infinite?

This isn't what they're saying. The misreading you've done here is analogous to the following:

"As $x \to 0$ in $(\cos x)/x$, the numerator remains finite, so the ratio becomes infinite."

"But why would nonzero $x$ cause the numerator to become infinite?"

I don't understand this part. Furthermore, why does this then imply that the volume density becomes infinite?

The point is that Gauss's law tells us that $$\int \text{div} \, \mathbf{D} \, dV = 4 \pi Q$$ where $Q$ is the charge inside the volume. On the other hand, the left-hand side is equal to $$\int \mathbf{D} \cdot \mathbf{n} \, dS.$$ As the surfaces shrink together, this quantity remains finite, in the sense that it doesn't go to zero. But this quantity is also equal to $4 \pi Q$, so the charge $Q$ remains finite (i.e. not zero). But to have a finite charge in an infinitesimal volume implies the volume charge density is infinite.